Theorem: Let f:\mathbb{R}\to\mathbb{C} be a function of rapid decay. Then \sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty \hat{f}(n), where \hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\; dx is the Fourier transform of f.

Proof: Write g(x)=\sum_{n=-\infty}^\infty f(x+n). The Fourier coefficients of g are a_n=\int_0^1\sum_{m=-\infty}^\infty f(x+m)e^{-2\pi inx}\; dx=\int_{-\infty}^\infty f(x)e^{-2\pi inx}\; dx=\hat{f}(n). Therefore the Fourier expansion of g is g(x)=\sum_{n=-\infty}^\infty \hat{f}(n)e^{2\pi inx}. Substituting x=0 gives \sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty \hat{f}(n), as desired.